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$ A B $ and $ C D $ are common tangents to two circles of equal radii. Prove that $ A B=C D $.
Given:
\( A B \) and \( C D \) are common tangents to two circles of equal radii.
To do:
We have to prove that \( A B=C D \).
Solution:
Join $OA, OC, OB$ and $OD$.
From the figure,
$\angle OAB = 90^o$ [Tangent at any point of a circle is perpendicular to the radius through the point of contact)
This implies,
$AC$ is a straight line.
$\angle OAB + \angle OCD = 180^o$
$AB\ \parallel\ CD$
Similarly,
$BD$ is a straight line and $\angle OBA = ∠ODC = 90^o$
$AC = BD$ (Radii of two circles are equal)
In quadrilateral $ABCD$,
$\angle A = \angle B = \angle C = \angle D = 90^o$
$AC = BD$
$ABCD$ is a rectangle.
Therefore,
$AB = CD$ (Opposite sides of a rectangle are equal)
Hence proved.
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